Find Asymptotes, Critical, and Inflection Points

This example describes how to analyze a simple function to find its asymptotes, maximum, minimum, and inflection point.

Define a Function

The function in this example is

$\mathit{f}\left(\mathit{x}\right)=\frac{3{\mathit{x}}^{2}+6\mathit{x}-1}{{\mathit{x}}^{2}+\mathit{x}-3}.$

First, create the function.

syms x
num = 3*x^2 + 6*x -1;
denom = x^2 + x - 3;
f = num/denom
f =

$\frac{3 {x}^{2}+6 x-1}{{x}^{2}+x-3}$

Plot the function by using fplot. The fplot function automatically shows horizontal and vertical asymptotes.

fplot(f)

Find Asymptotes

To find the horizontal asymptote of $f$ mathematically, take the limit of $f$ as $x$ approaches positive infinity.

limit(f,Inf)
ans = $3$

The limit as $x$ approaches negative infinity is also 3. This result means the line $y=3$ is a horizontal asymptote to $f$.

To find the vertical asymptotes of $f$, set the denominator equal to 0 and solve it.

roots = solve(denom)
roots =

$\left(\begin{array}{c}-\frac{\sqrt{13}}{2}-\frac{1}{2}\\ \frac{\sqrt{13}}{2}-\frac{1}{2}\end{array}\right)$

roots indicates that the vertical asymptotes are the lines

$\mathit{x}=\frac{-1-\sqrt{13}}{2}$

and

$\mathit{x}=\frac{-1+\sqrt{13}}{2}$.

Find Maximum and Minimum

You can see from the graph that $f$ has a local maximum between the points $x=–2$ and $x=0$. It also has a local minimum between $x=–6$ and $x=–2$. To find the $x$-coordinates of the maximum and minimum, first take the derivative of $f$.

f1 = diff(f)
f1 =

$\frac{6 x+6}{{x}^{2}+x-3}-\frac{\left(2 x+1\right) \left(3 {x}^{2}+6 x-1\right)}{{\left({x}^{2}+x-3\right)}^{2}}$

To simplify this expression, enter the following.

f1 = simplify(f1)
f1 =

$-\frac{3 {x}^{2}+16 x+17}{{\left({x}^{2}+x-3\right)}^{2}}$

Next, set the derivative equal to 0 and solve for the critical points.

crit_pts = solve(f1)
crit_pts =

$\left(\begin{array}{c}-\frac{\sqrt{13}}{3}-\frac{8}{3}\\ \frac{\sqrt{13}}{3}-\frac{8}{3}\end{array}\right)$

As the graph of $f$ shows, the function has a local minimum at

${\mathit{x}}_{1}=\frac{-8-\sqrt{13}}{3}$

and a local maximum at

${\mathit{x}}_{1}=\frac{-8+\sqrt{13}}{3}$.

Plot the maximum and minimum of f.

fplot(f)
hold on
plot(double(crit_pts), double(subs(f,crit_pts)),'ro')
title('Maximum and Minimum of f')
text(-4.8,5.5,'Local minimum')
text(-2,4,'Local maximum')
hold off

Find Inflection Point

To find the inflection point of $f$, set the second derivative equal to 0 and solve for this condition.

f2 = diff(f1);
inflec_pt = solve(f2,'MaxDegree',3);
double(inflec_pt)
ans = 3×1 complex

-5.2635 + 0.0000i
-1.3682 - 0.8511i
-1.3682 + 0.8511i

In this example, only the first element is a real number, so this is the only inflection point. MATLAB® does not always return the roots to an equation in the same order.

Instead of selecting the real root by indexing into inter_pt, identify the real root by determining which roots have a zero-valued imaginary part.

idx = imag(double(inflec_pt)) == 0;
inflec_pt = inflec_pt(idx)
inflec_pt =

$-\frac{13}{9 {\left(\frac{169}{54}-\frac{\sqrt{2197}}{18}\right)}^{1/3}}-{\left(\frac{169}{54}-\frac{\sqrt{2197}}{18}\right)}^{1/3}-\frac{8}{3}$

Plot the inflection point. The extra argument [-9 6] in fplot extends the range of $x$ values in the plot so that you can see the inflection point more clearly, as the figure shows.

fplot(f,[-9 6])
hold on
plot(double(inflec_pt), double(subs(f,inflec_pt)),'ro')
title('Inflection Point of f')
text(-7,1,'Inflection point')
hold off

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